Perfect fluid and Cauchy momentum equation

Question

The stress-energy tensor of a perfect fluid is given by

$$T^{\mu\nu}=\left(\rho+pc^{-2}\right)u^\mu u^\nu+pg^{\mu\nu}$$

The divergence of the stress-energy tensor is zero: $\nabla_\mu T^{\mu\nu}=0$. Hence

$$\nabla_\mu\left(\rho+pc^{-2}\right)u^\mu u^\nu+\nabla_\mu pg^{\mu\nu}=0$$

Expanding the first term, and using the product rule on the second term, yields

$$\nabla_\mu\rho u^\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}+p\nabla_\mu g^{\mu\nu}=0$$

Using the product rule once again on the first term yields

$$\left(\nabla_\mu\rho u^\mu\right) u^\nu+\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}+p\nabla_\mu g^{\mu\nu}=0$$

By the continuity equation, $\nabla_\mu\rho u^\mu=0$. Hence

$$\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}+p\nabla_\mu g^{\mu\nu}=0$$

The divergence of the metric tensor is zero: $\nabla_\mu g^{\mu\nu}=0$. Hence

$$\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\left(\nabla_\mu p\right)g^{\mu\nu}=0$$

Finally, using tensor contraction on the last term yields

$$\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\nabla^{\nu}p=0$$

We now turn to the Cauchy momentum equation in the Euler equations:

$$0=\rho\left(\frac{\partial}{\partial t}+\vec{u}\cdot\vec{\nabla}\right)\vec{u}+\vec{\nabla}p= \rho\left(c\nabla_0+\vec{u}\cdot\vec{\nabla}\right)\vec{u}+\vec{\nabla}p$$

Using the non-relativistic approximation $\gamma\approx1$ we obtain:

$$0\approx \rho\left(\gamma c\nabla_0+\gamma\vec{u}\cdot\vec{\nabla}\right)\gamma\vec{u}+\vec{\nabla}p= \rho u^{\mu}\nabla_{\mu}u^i+\nabla^ip$$

Compare this with the result obtained from the stress-energy tensor:

$$0=\rho u^\mu\nabla_\mu u^\nu+\nabla_\mu pc^{-2}u^\mu u^\nu+\nabla^{\nu}p$$

Why is there an extra term ($\nabla_\mu pc^{-2}u^\mu u^\nu$)? Does it vanish in the non-relativistic limit, simply because of the $c^{-2}$ factor?

Answer 1

Note that $\nabla_\mu (\rho u^\mu)=0$ is not correct (and not the continuity equation). Keep in mind that (based on your definition of the stress tensor) $\rho$ is the energy density, and the conserved energy current is $T_{0\mu}$.

The relativistic Euler (momentum conservation) equation is $$ D u_\mu = -\frac{1}{\rho+P}\nabla_\mu^\perp P $$ where $D=u^\mu\nabla_\mu$ and $\nabla_\mu^\perp=(g_{\mu\nu}-u_\mu u_\nu)\nabla^\nu$, which shows that the relativistic inertia is the enthalpy density, $w=\rho+P$. In the non-relativistic limit, $u_0\simeq 1$, $D$ is the comoving derivative, $w$ is the mass density, and $\nabla_i^\perp\simeq\nabla_i$. This leads to the standard Euler equation.